In Cr^{2+}, Mn^{2+}, Fe^{2+}, and Ni^{2+},the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $H_2O$ is a weak field ligand. The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)}

Vedclass · Accepted Answer

$[Ni(H_2O)_6]^{2+}$

Vedclass · Answer

(B) $H_2O$ is a weak field ligand. The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.For $[Cr(H_2O)_6]^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.For $[Mn(H_2O)_6]^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.For $[Fe(H_2O)_6]^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.For $[Ni(H_2O)_6]^{2+}$ $(3d^8)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.Since $[Ni(H_2O)_6]^{2+}$ has the lowest number of unpaired electrons $(n=2)$,it has the lowest magnetic moment.

In $Cr^{2+}, Mn^{2+}, Fe^{2+},$ and $Ni^{2+}$,the number of electrons in the $3d$ orbital are $3d^4, 3d^5, 3d^6,$ and $3d^8$ respectively. Which of the following aqueous complexes will have the lowest magnetic moment?

Similar Questions

The number of possible isomers of an octahedral complex $[Co(C_2O_4)_2(NH_3)_2]^-$ is

What is the spin-only magnetic moment value $(BM)$ of a divalent metal ion with atomic number $25$ in its aqueous solution?

What is the magnetic moment of a divalent ion in aqueous solution if its atomic number is $28$ (in $BM$)?

$A$ chemist wants to determine the molecular geometry of the $[CoCl_4]^{2-}$ ion. Which of the following gives the best suggestion for a measurement and for the interpretation of that measurement?

The number of possible isomers for the square planar complex $K_2[PdClBr_2SCN]$ is .................

Vedclass Test Series

Exam Paper Generator

Online Exam Module